30
Jul 19

Action of a matrix in its root subspace

Action of a matrix in its root subspace

The purpose of the following discussion is to reveal the matrix form of A in N_{\lambda }^{(p)}.

Definition 1. Nonzero elements of N_{\lambda }^{(p)} are called root vectors. This definition can be detailed as follows:

Elements of N_{\lambda }^{(1)}\setminus \{0\} are eigenvalues.

Elements of N_{\lambda }^{(2)}\setminus N_{\lambda }^{(1)} are called root vectors of 1st order.

...

Elements of N_{\lambda }^{(p)}\setminus N_{\lambda }^{(p-1)} are called root vectors of order p-1.

Thus, root vectors belong to

N_{\lambda }^{(p)}\setminus \{0\}=\left( N_{\lambda }^{(p)}\setminus  N_{\lambda }^{(p-1)}\right) \cup ...\cup \left( N_{\lambda }^{(2)}\setminus  N_{\lambda }^{(1)}\right) \cup \left( N_{\lambda }^{(1)}\setminus \{0\}\right)

where the sets of root vectors of different orders do not intersect.

Exercise 1. (A-\lambda I)\left( N_{\lambda }^{(k)}\setminus N_{\lambda }^{(k-1)}\right) \subset \left( N_{\lambda }^{(k-1)}\setminus N_{\lambda}^{(k-2)}\right) .

Proof. Suppose x\in N_{\lambda }^{(k)}\setminus N_{\lambda }^{(k-1)}, that is, (A-\lambda I)^{k}x=0 and (A-\lambda I)^{k-1}x\neq 0. Denoting y=(A-\lambda I)x, we have (A-\lambda I)^{k-1}y=0 and (A-\lambda I)^{k-2}y\neq 0, which means that y\in N_{\lambda }^{(k-1)}\setminus N_{\lambda }^{(k-2)} and A-\lambda I maps N_{\lambda }^{(k)}\setminus N_{\lambda }^{(k-1)} into N_{\lambda }^{(k-1)}\setminus N_{\lambda}^{(k-2)}.

Now, starting from some x_{p}\in N_{\lambda }^{(p)}\setminus N_{\lambda}^{(p-1)}, we extend a chain of root vectors all the way to an eigenvector. By Exercise 1, the vector x_{p-1}=(A-\lambda I)x_{p} belongs to N_{\lambda }^{(p-1)}\setminus N_{\lambda }^{(p-2)}. From the definition of x_{p-1} we see that

(1) Ax_{p}=\lambda x_{p}+x_{p-1},   x_{p-1}\in N_{\lambda}^{(p-1)}\setminus N_{\lambda }^{(p-2)}

(x_{p} is an "eigenvector" up to a root vector of lower order). Similarly, denoting x_{p-2}=(A-\lambda I)x_{p-1}, we have

(2) Ax_{p-1}=\lambda x_{p-1}+x_{p-2},   x_{p-2}\in N_{\lambda}^{(p-2)}\setminus N_{\lambda }^{(p-3)}.

...

Continuing in the same way, we get x_{1}=(A-\lambda I)x_{2}\in N_{\lambda}^{(1)}\setminus \{0\},

(3) Ax_{2}=\lambda x_{2}+x_{1},   x_{1}\in N_{\lambda}^{(1)}\setminus \{0\},

(4) Ax_{1}=\lambda x_{1},   x_{1}\neq 0.

Exercise 2. The vectors x_{1},...,x_{p} defined above are linearly independent.

Proof. If \sum_{j=1}^{p}a_{j}x_{j}=0, then a_{p}x_{p}=-\sum_{j=1}^{p-1}a_{j}x_{j}. Here the left side belongs to N_{\lambda}^{(p)}\setminus N_{\lambda }^{(p-1)} and the right side belongs to N_{\lambda }^{(p-1)} because of inclusion relations. Hence, a_{p}=0. Similarly, all other coefficients are zero.

By Exercise 2, the vectors x_{1},...,x_{p} form a basis in L=span(x_{1},...,x_{p}).

Exercise 3. The transformation A in L is given by the matrix

(5) A=\left(\begin{array}{ccccc}\lambda & 1 & 0 & ... & 0 \\0 & \lambda & 1 & ... & 0 \\  ... & ... & ... & ... & ... \\0 & 0 & 0 & ... & 1 \\0 & 0 & 0 & ... & \lambda\end{array}  \right) =\lambda I+J

where J is a matrix with unities in the first superdiagonal and zeros everywhere else.

Proof. Since x_{1},...,x_{p} is taken as the basis, x_{i} can be identified with the unit column-vector e_{i}. The equations (1)-(4) take the form

Ae_{1}=\lambda e_{1}, Ae_{j}=\lambda e_{j}+e_{j-1}, j=2,...,p.

Putting these equations side by side we get

AI=A\left( e_{1},...,e_{p}\right) =\left( Ae_{1},...,Ae_{p}\right) =\left(\lambda e_{1},\lambda e_{2}+e_{1},...,\lambda e_{p}+e_{p-1}\right) =

=\left(\begin{array}{cccc}\lambda & 1 & ... & 0 \\0 & \lambda & ... & 0 \\  ... & ... & ... & ... \\0 & 0 & ... & 1 \\0 & 0 & ... & \lambda\end{array}\right) .

Definition 2. The matrix in (5) is called a Jordan cell.

Leave a Reply

You must be logged in to post a comment.